w w ap eP m e tr .X w UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS 9702 PHYSICS 9702/02 Paper 2 (AS Structured Questions), maximum raw mark 60 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. om .c MARK SCHEME for the October/November 2007 question paper s er GCE Advanced Subsidiary Level and GCE Advanced Level Page 2 1 2 Mark Scheme GCE A/AS LEVEL – October/November 2007 Syllabus 9702 (a) systematic: e.g. constant error (in all readings) cannot be eliminated by averaging error in measuring instrument random: e.g. readings scattered (equally) about true value error due to observer can be eliminated by averaging (only if averaging not included for systematic) B1 [2] (b) 15 = π × R2 × 20 R = 0.4886 cm (accept any number of s.f.) % uncertainty in V = 3.3 % (or 0.5/15) % uncertainty in L = 0.5 % (or 0.1/20) % uncertainty in R = 1.9 % (i.e. one half of the sum) R = 0.489 ± 0.009 cm C1 C1 C1 C1 A1 [5] (a) 3.5 T B1 [1] (b) (i) distance = average speed × time (however expressed) = 14 m C1 A1 [2] A1 [1] (c) 3.5T = 14 + 5.6(T – 5) T = 6.7 s C1 A1 [2] (d) (i) acceleration = (5.6 / 5 =) 1.12 m s–2 force = ma = 75 N C1 C1 A1 [3] C1 A1 [2] B1 [1] B1 B1 [2] C1 A1 [2] C1 A1 [2] (ii) distance = 5.6 × (T – 5) (or 3.5T – 14) (ii) power = (force × speed =) {75 + 23} × 4.5 = 440 W (allow 1/2 for 234 W, 0/2 for 338 W or 104 W) 3 Paper 02 (a) (i) potential energy: stored energy available to do work (ii) gravitational: due to height/position of mass OR distance from mass OR moving mass from one point to another elastic: due to deformation/stretching/compressing (b) (i) height raised = (61 – {61 cos18} =) 3.0 cm energy = (mgh = 0.051 × 9.8 × 0.030 =) 1.5 × 10–2 J (ii) moment = force × perpendicular distance = 0.051 × 9.8 × 0.61 × sin18 = 0.094 N m © UCLES 2007 B1 Page 3 4 5 Mark Scheme GCE A/AS LEVEL – October/November 2007 Paper 02 (a) brittle B1 (b) Young modulus = stress / strain = (9.5 × 108) / 0.013 = 7.3 × 1010 Pa (allow ± 0.1 × 1010 Pa) C1 (c) stress = force / area (minimum) area = (1.9 × 103) / (9.5 × 108) = 2.0 × 10–6 m2 (max) area of cross-section = (3.2 – 2.0) × 10–6 = 1.2 × 10–6 m2 C1 C1 A1 [3] (d) when bent, ‘top’ and ‘bottom’ edges have different extensions with thick rod, difference is greater (than with a thin rod) so breaks with less bending M1 A1 A0 [2] (a) amplitude between 6.5 squares and 7.5 squares on 3 peaks (allow 1 mark if outside this range but between 6.0 and 8.0 squares) correct phase (ignore lead/lag, look at x-axis only and allow ±½ square B2 B1 [3] (b) λ = ax / D 540 × 10–9 = (0.700 × 10–3 x) / 2.75 x = 2.12 mm C1 C1 A1 [3] (c) (i) same separation bright areas brighter (1) dark areas, no change (1) (allow ‘contrast greater’ for 1 mark if dark/light areas not discussed) fewer fringes observed (1) any two, 1 each B1 A1 [1] [2] B2 [3] B1 B1 [2] (a) power = VI current = 10.5 × 103 / 230 = 45.7 A C1 M1 A0 [2] (b) (i) p.d. across cable = 5.0 V R = 5.0 / 46 = 0.11 Ω C1 C1 A1 [3] C1 C1 A1 [3] (ii) smaller separation of fringes no change in brightness 6 Syllabus 9702 (ii) R = ρL / A 0.11 = (1.8 × 10–8 × 16 × 2) / A A = 5.3 × 10–6 m2 (wires in parallel, not series, allow max 1/3 marks) © UCLES 2007 Page 4 Mark Scheme GCE A/AS LEVEL – October/November 2007 (c) (i) either power = V 2 / R or power ∝ V 2 ratio = (210 / 230)2 = 0.83 (ii) resistance of cable is greater greater power loss/fire hazard/insulation may melt wire may melt/cable gets hot 7 Syllabus 9702 Paper 02 C1 A1 [2] M1 A1 [2] (a) most α-particles deviated through small angles (accept ‘undeviated’) few α-particles deviated through angles greater than 90° B1 B1 [2] (b) (i) allow 10–9 m → 10–11 m B1 [1] B1 [1] (ii) allow 10–13 m → 10–15 m (if (i) and (ii) out of range but (ii) = 10–4(i), then allow 1 mark) (if no units or wrong units but (ii) = 10–4(i), then allow 1 mark) © UCLES 2007